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Mathematics 12 Online
OpenStudy (anonymous):

3. Christy drove 300 miles on her vacation. She drove an average of 1.25 times faster on the second 150 miles of her trip than she did on the first 150 miles of her trip. Which expression represents the time she spent driving? Let x = her speed on the first half of the trip

OpenStudy (anonymous):

375/x 337.5/x 270x 270/x

OpenStudy (yanasidlinskiy):

Are those your answers?? Or the mutltiple choices??

OpenStudy (anonymous):

multiple choices

OpenStudy (paki):

these are the MQs ....

OpenStudy (yanasidlinskiy):

Ok:) Here we go....

OpenStudy (yanasidlinskiy):

During the first 150 miles, she drove 150 miles at a speed of \(x\) miles per hour, so her time was \([150 \text{ miles} = x * t_{1st}]where (t_{1st}\) is the time she spent driving the first 150 miles. On the second half of the trip, she drove at \(1.25x\) miles per hour. There, the equation would be \[150 \text{ miles} = 1.25x * t_{2nd}\]where \(t_{2nd}\) is the time she spent driving the second 150 miles. Solve both of those equations for the \(t\) variables, then add them together. Your answer will be \(t_{1st} + t_{2nd}\)

OpenStudy (paki):

i think it's C... @YanaSidlinskiy

OpenStudy (yanasidlinskiy):

Remember the distance formula: \(d = rt\) where \(d\) is distance traveled, \(r\) is speed or rate, and \(t\) is time!:) I think so too:)

OpenStudy (anonymous):

thanks guys!

OpenStudy (yanasidlinskiy):

You're Welcome!!!!:D

OpenStudy (paki):

yeah formula is s=v/t.... or t=s*v....

OpenStudy (paki):

pleasure ....

OpenStudy (yanasidlinskiy):

sure:)

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